Help Understanding Explicit Automorphism Formulas for Semi-Direct Products

Question

I'm currently working through some material on semi-direct products, and I'm totally confused by some of the explicit examples given for semi-direct products, namely how one determines the automorphisms.

For example, In classifying groups of order $30$, done here specifically, the answer makes sense in that there are four possible automorphisms given by $\varphi(1) \in \{(0,0), (1,0), (0,2), (1,2)\}$, but what are these automorphisms?

For $\varphi(1) = (0,0)$, I get this is a trivial map corresponding to simply the direct product.

What about $\varphi(1) = (1,0)$, what's an explicit formula for this automorphism? Semi-direct products of this form arise from the action of conjugation of the right factor $K$ on the left factor $H$ (for a semi direct product $H \rtimes_{\varphi} K$). In this case, we have a Sylow $2$ subgroup $P_2 \cong \mathbb{Z}_2$, and a group $N \cong \mathbb{Z}_{15}$ of order $15$.

The theory states that if $\varphi: K \longrightarrow \text{Aut}(H)$, then $\varphi(k)[(h)] = khk^{-1}$. So if $\varphi(1) = (1,0)$, then wouldn't we simply have for $n \in \mathbb{Z}_{15}$: $$\varphi(1)(n) = 1+n-1 = n?$$ This is the trivial automorphism however, which is in direct contradiction to the fact that $\varphi(1) = (1,0)$ isn't trivial.

Essentially what I'm asking for is an explanation on how one deduces that (in the context of the linked post): "if $\varphi(1)=(1,0)$, then $P_2 \rtimes_{\varphi} N \cong D_{15}$ because _____, and if $\varphi(1)=(0,2)$, then $P_2 \rtimes_{\varphi} N \cong \mathbb{Z}_5 \times S_3$ because ___..."

Could someone explain what's going on here?

Answer 1

Groundwork

We are trying to understand semidirect products of the form $ℤ_{15} ⋊_φ ℤ_2$. We rely on the following observations:

• Let $R$ be a ring and let $u$ be an element of $R^×$. The map $$ λ_u \colon R \to R \,, \quad x \mapsto ux $$ is an automorphism of the underlying additive group of $R$. We have thus a homomorphism of groups $$ λ \colon R^× \to \operatorname{Aut}_ℤ(R) \,, \quad u \mapsto λ_u \,. $$ This homomorphism is always injective because $u$ can be retrieved from $λ_u$ as $u = λ_u(1)$. If $R = ℤ_n$, then $λ$ is also surjectiv, and thus an isomorphism of groups $\operatorname{Aut}_ℤ(ℤ_n) ≅ ℤ_n^×$.

• Given two coprime integers $m$ and $n$, we have an isomorphism of rings (and thus in particular of abelian gorups) $ℤ_{mn} ≅ ℤ_m × ℤ_n$ given by $[k] \mapsto ([k], [k])$.

• For every two rings $R$ and $S$, we have $(R × S)^× = R^× × S^×$.

• Given any group $G$, homomorphism of groups from $ℤ_2$ is $G$ are uniquely determined by their image of the element $1$ of $ℤ_2$. This image can be any element of $G$ whose order divides $2$. (In other words, this element can either be the neutral element of $G$, or an element of order $2$. The neutral element corresponds to the trivial homomorphism from $ℤ_2$ to $G$.)

Let us fix a piece of notation:

• Let $R$ be a ring and let $u$ be an element of $R^×$ whose order divides $2$. Let $ψ_u$ be the resulting homomorphism of groups from $ℤ_2$ to $\operatorname{Aut}_ℤ(R)$ with $ψ_u(1) = λ_u$. We then abbreviate the resulting semidirect product $R ⋊_{ψ_u} ℤ_2$ as $R ⋊_u ℤ_2$. (Inside of the group $R ⋊_u ℤ_2$, the conjucation action of $ℤ_2$ on $R$ is thus given by multiplication with $u$.)

We will need one very explicit example of this general construction:

• For every positive integer $n$, we have $ℤ_n ⋊_{-1} ℤ_2 ≅ \mathrm{D}_n$. (We use here the standard description of the dihedral group $\mathrm{D}_n$ as a semidirect product of $ℤ_n$ by $ℤ_2$.)

We will also need the following observation:

• Let $R$ and $S$ be two rings.
  • Let $r$ be an element of $R^×$ whose order divides $2$. We then get the element $(r, 1)$ of $(R × S)^×$, whose order again divides $2$. The resulting action of $ℤ_2$ on $R × S$ does nothing on the second factor. (More explicitly, the element $1$ of $ℤ_2$ acts on the first factor by multiplication with $r$, and on the second factor by multiplication with $1$.) We get from this observation an isomorphism of groups $$ (R × S) ⋊_{(r, 1)} ℤ_2 ≅ (R ⋊_R ℤ_2) × S \,. $$
  • Similarly, if $s$ is an element of $S^×$ whose order divides $2$, then $$ (R × S) ⋊_{(1, s)} ℤ_2 ≅ R × (S ⋊_s ℤ_2) \,. $$

Computation of elements

Let us examine the chain of isomorphisms $$ ℤ_{15}^× ≅ (ℤ_3 × ℤ_5)^× = ℤ_3^× × ℤ_5^× ≅ ℤ_2 × ℤ_4 $$ in more detail:

• The isomorphism $ℤ_{15} ≅ ℤ_3 × ℤ_5$ comes from the chinese reminder theorem because $3$ and $5$ are coprime. More explicitly, the map $$ θ \colon ℤ_{15} \to ℤ_3 × ℤ_5 \,, \quad [n] \mapsto ([n], [n]) $$ is an isomorphism. We can also give an explicit formula for the inveres of $θ$: we have $θ^{-1}(1, 0) = 10 = -5$ and $θ^{-1}(0, 1) = 6$, and therefore $$ θ^{-1}([m], [n]) = [-5m + 6n] \,. $$

• The groups $ℤ_3^×$ and $ℤ_5^×$ are cyclic, and generated by the elements $2$ and $2$ respectively. We can therefore choose the isomorphism $ℤ_3^× × ℤ_5^× ≅ ℤ_2 × ℤ_4$ so that an element $([k], [l])$ of $ℤ_2 × ℤ_4$ corresponds to the element $(2^k, 2^l)$ of $ℤ_3^× × ℤ_5^×$.

For an element $([k], [l])$ of $ℤ_2 × ℤ_4$ we therefore find that

• the resulting element of $ℤ_3^× × ℤ_5^×$ is given by $(2^k, 2^l)$;
• the resulting element of $ℤ_{15}^×$ is given by $[-5 ⋅ 2^k + 6 ⋅ 2^l]$.

We have now the following elements in the groups $ℤ_2 × ℤ_4$, $ℤ_3^× × ℤ_5^×$ and $ℤ_{15}^×$ whose order divides $2$.

$ℤ_2 × ℤ_4 $	$ℤ_3^× × ℤ_5^×$	$ℤ_{15}^×$
$(0, 0)$	$(1, 1)$	$-5 + 6 = 1$
$(0, 2)$	$(1, 4) = ( 1, -1)$	$-5 + 24 = 19 = 4$
$(1, 0)$	$(2, 1) = (-1, 1)$	$-10 + 6 = -4$
$(1, 2)$	$(2, 4) = (-1, -1)$	$-10 + 24 = 14 = -1$

For every element of $ℤ_2 × ℤ_4$, the corresponding automorphisms of $ℤ_{15}$ are simply given by multiplication with corresponding element of $ℤ_{15}^×$. As an example, the automorphism of $ℤ_{15}$ corresponding to the element $(1, 0)$ of $ℤ_4 × ℤ_2$ is given by multiplication with $-4$.

Computation of semidirect products

Let us now compute the resulting semidirect products.

• The element $(0, 0)$ of $ℤ_2 × ℤ_4$ corresponds to the element $1$ of $ℤ_{15}^×$. The resulting automorphism of $ℤ_{15}$ is simply the identity automorphism, and the resulting homomorphism from $ℤ_2$ to $\operatorname{Aut}(ℤ_{15})$ is the trivial homomorphism. And the semidirect product resulting from the trivial homomorphism is simpli the direct product. Therefore, $$ ℤ_{15} ⋊_1 ℤ_2 ≅ ℤ_{15} × ℤ_2 ≅ ℤ_{30} \,. $$

• The element $(0, 2)$ of $ℤ_2 × ℤ_4$ corresponds to the element $4$ of $ℤ_{15}$. To compute the resulting semidirect product $ℤ_{15} ⋊_4 ℤ_2$, we employ once again the isomorphism $ℤ_{15} ≅ ℤ_3 × ℤ_5$. The corresponding element in $ℤ_3 × ℤ_5$ is given by $(1, -1)$. We have therefore an isomorphism of groups $$ ℤ_{15} ⋊_4 ℤ_2 ≅ (ℤ_3 × ℤ_5) ⋊_{(1, -1)} ℤ_2 ≅ ℤ_3 × (ℤ_5⋊_{-1} ℤ_2) ≅ ℤ_3 × \mathrm{D}_5 \,. $$

• For the element $(1, 0)$ of $ℤ_2 × ℤ_4$ we find similarly that $$ ℤ_{15} ⋊_{-4} ℤ_2 ≅ (ℤ_3 × ℤ_5) ⋊_{(-1, 1)} ℤ_2 ≅ (ℤ_3 ⋊_{-1} ℤ_2) × ℤ_5 ≅ \mathrm{D}_3 × ℤ_5 ≅ \mathrm{S}_3 × ℤ_5 \,. $$

• For the element $(1, 2)$ of $ℤ_2 × ℤ_4$ the corresponding element of $ℤ_{15}^×$ is given by $-1$. We find that $$ ℤ_{15} ⋊_{-1} ℤ_2 ≅ \mathrm{D}_{15} \,. $$

Answer 2

My answer is shorter.

If I understand you correctly, first we need to figure out what the automorphisms of the group $\mathbb{Z}_{15}$ look like.

Let $A=\langle5\rangle$, $B=\langle3\rangle$. We have $\mathbb{Z}_{15}=A\oplus B$ and $|A|=3$, $|B|=5$. Since $A$ and $B$ have coprime orders and $\operatorname{Aut}(A)\cong\mathbb{Z}_2$, $\operatorname{Aut}(B)\cong\mathbb{Z}_4$, it follows that $$ \operatorname{Aut}(\mathbb{Z}_{15}) \cong\operatorname{Aut}(A)\times\operatorname{Aut}(B) \cong\mathbb{Z}_2\times\mathbb{Z}_4. $$

Therefore, it is natural to assume that
the automorphism $\alpha=(1,0)$ acts as follows $\alpha(a)=-a$ for all $a\in A$ and $\alpha(b)=b$ for all $b\in B$ or $$ \alpha(5q)=-5q,\ \alpha(3p)=3p,\ q=1,2;\ p=1,2,3,4. $$ It can be given by a single formula $\alpha(1)=11$ (Check).

This means that $\mathbb{Z}_{15}\rtimes_{\varphi_1} K\cong D_3\times \mathbb{Z}_5$, where $\varphi_1(t)=\alpha$ (we assume that $|K|=2$ and $K=\langle t\rangle$).

Similarly $\beta=(0,2)$ acts as follows $\beta(a)=a$ for all $a\in A$ and $\beta(b)=-b$ for all $b\in B$ or $$ \beta(5q)=5q,\ \beta(3p)=-3p,\ q=1,2;\ p=1,2,3,4; $$ or $\beta(1)=4$ (Check).

If $\varphi_2(t)=\beta$, then $\mathbb{Z}_{15}\rtimes_{\varphi_2} K\cong \mathbb{Z}_3\times D_5$.

And the last $\gamma=(1,2)$ acts as follows $\gamma(a)=-a$ for all $a\in A$ and $\gamma(b)=-b$ for all $b\in B$ or $\gamma(1)=-1$.

If $\varphi_3(t)=\gamma$, then $\mathbb{Z}_{15}\rtimes_{\varphi_3} K\cong D_{15}$.

Answer 3

The setting was $G \cong \mathbb Z/15\mathbb Z \rtimes_{\varphi} \mathbb Z/2\mathbb Z$, where $\varphi : \mathbb Z/2\mathbb Z \to \operatorname{Aut}(\mathbb Z/15\mathbb Z)$ is a group homomorphism.

1. Find an explicit description of $\varphi$

We have $$\operatorname{Aut}(\mathbb Z/15\mathbb Z) \cong (\mathbb Z/15\mathbb Z)^\times \cong (\mathbb Z/3\mathbb Z \times \mathbb Z/5\mathbb Z)^\times = (\mathbb Z/3\mathbb Z)^\times \times (\mathbb Z/5\mathbb Z)^\times \cong \mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z,$$ such that we may (for the moment) think of $\varphi$ as a map $\mathbb Z/2\mathbb Z \to\mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z$

Clearly, $\varphi$ is determined by the image $\varphi(1)$. Let us consider the example $\varphi(1) =(0,2) \in\mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z$. To get the image $\varphi(1)\in \operatorname{Aut}(\mathbb Z/15\mathbb Z)$ in the original setting, we need to trace back the element $(0,2)\in\mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z$ in the above chain of isomorphisms.

First step: $(\mathbb Z/3\mathbb Z)^\times \times (\mathbb Z/5\mathbb Z)^\times \cong \mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z$.

In $(0,2)\in\mathbb Z/2\mathbb Z \times \mathbb Z/4\mathbb Z$, the component $0$ is the neutral element in $\mathbb Z/2\mathbb Z$ and $2$ is the unique element of order $2$ in $\mathbb Z/4\mathbb Z$. Therefore, the preimage of the first component $0$ is the neutral element in $(\mathbb Z/3\mathbb Z)^\times$, which is $1$, and the preimage of the second component $1$ in $(\mathbb Z/5\mathbb Z)^\times$ is $-1$, which has order $2$. Together, the sought after preimage is $(1,-1)$.

Second step $(\mathbb Z/15\mathbb Z)^\times \cong (\mathbb Z/3\mathbb Z)^\times \times (\mathbb Z/5\mathbb Z)^\times$

This is based on the Chinese remainder theorem $\mathbb Z/15\mathbb Z \cong \mathbb Z/3\mathbb Z\times \mathbb Z/5\mathbb Z$. For the preimage of $(1,-1)$ we need an integer having remainder $1$ mod $3$ and remainder $-1$ mod $5$. There is an algorithm for this computation (based on the Euclidean algorithm), but in this small case we simply check a few numbers and find the preimage $4\in 15 \mathbb Z/\mathbb Z$.

Third step $\operatorname{Aut}(\mathbb Z/15\mathbb Z) \cong (\mathbb Z/15\mathbb Z)^\times$

This is easy. The concrete isomorphism (in the direction $\leftarrow$) is $a + \mathbb Z/15\mathbb Z \mapsto (x \mapsto ax)$. In our case, we get $x \mapsto 4x$

Fourth step We have found $\varphi(1) = (x \mapsto 4x)$. What is $\varphi(a)$ for general $a\in\mathbb Z/2\mathbb Z$. In this case, this is very easy, as the only other element of $\mathbb Z/2\mathbb Z$ is the neutral element $0$. Of cource, $\varphi(0) = \operatorname{id}_{\mathbb Z/15\mathbb Z}$.

For general $a\in\mathbb Z/2\mathbb Z$ we may combine this as as $$\varphi(a) = (x \mapsto 4^a x).$$

(Remark. This would also work for larger cyclic groups instead of $\mathbb Z/2\mathbb Z$ using $a = 1 + \ldots + 1$ and the homomorphism property of $\varphi$.)

2. Determine the multiplication rule in the semidirect product

The general multiplication rule in the outer semidirect product $N \rtimes_{\varphi} U$ is $$(n_1, u_1) \cdot (n_2, u_2) = (n_1\cdot \varphi(h_1)(n_2), h_1 \cdot h_2)$$

For elements $(a_1,b_1), (a_2,b_2)\in \mathbb Z/15\mathbb Z \times \mathbb Z/2\mathbb Z$ and $\varphi$ as above we get $$(a_1,b_1) \cdot (a_2,b_2) = (a_1 + 4^{b_1} a_2,\, b_1 + b_2)$$

If you understand the above, you should be able to find the explicit multiplication rule also in the remaining cases.

3. Determine the group

We want to find out to which of the candidates $D_{15}$, $\mathbb Z/5\mathbb Z\times S_3$ and $\mathbb Z/3\mathbb Z\times D_5$ our group $\mathbb Z/15\mathbb Z\rtimes \mathbb Z/2\mathbb Z$ is isomorphic.

In general, checking two groups for isomorphism can be pretty hard. But in our case, counting the elements of order $2$ should be sufficient to distinguish the groups.

We count the elements of $(a,b)\in\mathbb Z/15\mathbb Z\rtimes \mathbb Z/2\mathbb Z$ of order $2$. We consider the equation $(a,b)\cdot (a,b) = (0,0)$. The second component is $2b = 0$, which is always true. The first component is $a(1 + 4^b) = 0$. For $b = 0$ this reduces to $2a = 0$ with the only solution $a = 0$. For $b = 1$ we get $5a = 0$ with five solutions $a\in\{0,3,6,9,12\}$ possible. Discarding the neutral element $(0,0)$, we have found the following five elements of order 2: $$\{(0,1), (3,1), (6,1), (9,1), (12,1)\}$$

Now you can count the elements of order $2$ in $D_{15}$, $\mathbb Z/5\mathbb Z\times S_3$ and $\mathbb Z/3\mathbb Z\times D_5$ to find out which one the considered semidirect product ist. (BTW, this also shows that the 3 candidate groups are pairwise non-isomorphic, as their number of elements of order $2$ differs.)